document.write( "Question 978065: The tangent to the curve y=ax^2+bx+2 at (1,0.5) is parallel to the normal to the curve y=x^2+6x+4 at (-2,-4).Find the value of a and b. \n" ); document.write( "
Algebra.Com's Answer #599573 by Edwin McCravy(20056)\"\" \"About 
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The tangent to the curve y=ax^2+bx+2 at (1,0.5) is parallel to the normal to the
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document.write( "To prevent students from simply taking our answers here and submitting them for\r\n" );
document.write( "a grade and learning nothing from them, here is a problem exactly in every\r\n" );
document.write( "detail like the one you submitted but with different numbers and different\r\n" );
document.write( "answers.  But every step is explained:\r\n" );
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\n" ); document.write( "The tangent to the curve y=ax^2+bx+36.5 at (6,-2.5) is parallel to the normal to
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document.write( "We begin by drawing the graph of y=x^2-4x-4, with a tangent (in green) and a\r\n" );
document.write( "normal line (in blue) at (3,-7): \r\n" );
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document.write( "We need the slope of the green tangent line so that we can find the slope of the\r\n" );
document.write( "blue normal line by taking its reciprocal with the opposite sign.\r\n" );
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document.write( "The derivative IS an expression for the slope of the tangent line at any point,\r\n" );
document.write( "so we find the derivative of:\r\n" );
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document.write( " y = x^2-4x-4\r\n" );
document.write( "y' = 2x-4\r\n" );
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document.write( "and evaluate it at the point (3,-7), using only its x value x=3\r\n" );
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document.write( "y'@x=3 = 2(3)-4 = 6-4 = 2, the slope of the green tangent line.\r\n" );
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document.write( "That means that the slope of the blue normal line is its reciprocal with the\r\n" );
document.write( "opposite sign, or -1/2.\r\n" );
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document.write( "Now we know that the graph of y = ax^2+bx+36.5 goes through the point (6,-2.5),\r\n" );
document.write( "since it is tangent there.  So we draw a line parallel to the blue line through\r\n" );
document.write( "(6,-2.5) which will be tangent to the graph of y = ax^2+bx+36.5 at (6,-2.5).\r\n" );
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document.write( "And we can approximately sketch in the graph of y = ax^2+bx+36.5 so that it\r\n" );
document.write( "will be tangent to the upper blue line at (6,-2.5):\r\n" );
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document.write( "Now as in the other graph, we know that the derivative IS the slope of the\r\n" );
document.write( "tangent line, so we find the derivative of\r\n" );
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document.write( " y = ax^2+bx+36.5\r\n" );
document.write( "y' = 2ax+b\r\n" );
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document.write( "and evaluate it at the point (6,-2.5), using only its x value x=6\r\n" );
document.write( "y'@x=6 = 2a(6)+b = 12a+b, the slope of the tangent line to\r\n" );
document.write( "y = ax^2+b+36.5 at (6,-2.5) which is the slope of the upper blue tangent line.\r\n" );
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document.write( "So we set that slope 12a+b equal to -1/2\r\n" );
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document.write( "12a+b = -1/2, multiplying through by 2:\r\n" );
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document.write( "eq. (1)       24a+2b = -1\r\n" );
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document.write( "We also know that y = ax^2+b+36.5 goes through the point (6,-2.5) so we\r\n" );
document.write( "substitute x=6 and y=-2.5 into that equation\r\n" );
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document.write( "   y = ax^2+bx+36.5\r\n" );
document.write( "-2.5 = a(6)^2+b(6)+36.5\r\n" );
document.write( "-2.5 = a(36)+6b+36.5\r\n" );
document.write( "-2.5 = 36a+6b+36.5\r\n" );
document.write( " -39 = 36a+6b\r\n" );
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document.write( "Divide through by 3 and reverse left and right sides\r\n" );
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document.write( "eq. (2)     12a+2b=-13\r\n" );
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document.write( "Now we solve the system of equations eq. (1) and eq. (2)\r\n" );
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document.write( "\"system%2824a%2B2b+=+-1%2C12a%2B2b=-13%29\"\r\n" );
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document.write( "We multiply the second equation through by -1 and add to eliminate b:\r\n" );
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document.write( "\"system%2824a%2B2b+=+-1%2C-12a-2b=13%29\"\r\n" );
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document.write( "12a = 12\r\n" );
document.write( "  a = 1\r\n" );
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document.write( "Substituting in eq. (1)\r\n" );
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document.write( "24a+2b = -1\r\n" );
document.write( "24(1)+2b = -1\r\n" );
document.write( "  24+2b = -1\r\n" );
document.write( "      2b = -25\r\n" );
document.write( "       b = -25/2\r\n" );
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document.write( "Answer:  a = 1,  b = -25/2 = -12.5\r\n" );
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document.write( "[So the equation of the green graph is y = x^2-12.5x+36.5]\r\n" );
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document.write( "Now do your problem, which is exactly in every detail like this one.\r\n" );
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document.write( "Edwin
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